Problem: A few families took a trip to an amusement park together. Tickets cost $$7.50$ each for adults and $$3.50$ each for kids, and the group paid $$40.00$ in total. There were $2$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${7.5x+3.5y = 40}$ ${x = y-2}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-2}$ for $x$ in the first equation. ${7.5}{(y-2)}{+ 3.5y = 40}$ Simplify and solve for $y$ $ 7.5y-15 + 3.5y = 40 $ $ 11y-15 = 40 $ $ 11y = 55 $ $ y = \dfrac{55}{11} $ ${y = 5}$ Now that you know ${y = 5}$ , plug it back into ${x = y-2}$ to find $x$ ${x = }{(5)}{ - 2}$ ${x = 3}$ You can also plug ${y = 5}$ into ${7.5x+3.5y = 40}$ and get the same answer for $x$ ${7.5x + 3.5}{(5)}{= 40}$ ${x = 3}$ There were $3$ adults and $5$ kids.